# That elevator analogy

1. May 1, 2013

### Paintjunkie

So If I remember right. The explanation as to why people in orbit appear to be weightless is cause they are in free fall. and because they are in an innertail frame we see them as weightless.

this explanation bothers me some. cause I feel as though that innertail frame is still expieriencing gravity. if that inertial frame was no where near earth or any other large mass wouldnt we be able to tell the difference.

I guess I am thinking that since I am made of mass I have a gravitational pull. so if i let go of an apple it would orbit me, if i was no where near a larger body. but if I was on the space station and let go of an apple it would just fall with me. Am i thinking about this wrong?

2. May 1, 2013

### Travis_King

An apple wouldn't orbit you under any real, obtainable conditions.

3. May 1, 2013

### Staff: Mentor

You do indeed exert a gravitational force on objects near you like the apple. However, in these elevator/free-fall thought experiments we ignore your gravity, and instead assume that the only gravitational attraction that matters is that of the earth.

And as for why that's a valid assumption? Try calculating the gravitational force between you and a nearby apple using $F=\frac{Gm_{1}m_{2}}{r^2}$ where r is in meters, the two masses are kilograms, and G is about 6x10-11.... You're no more likely to notice it in outer space than you do here on earth.

4. May 1, 2013

### Paintjunkie

Ok, i get that the force between us would be almost nothing. but in a universe with only me and the apple wouldnt almost nothing, be alot? wouldnt the apple and I be attracted to each other over a really long time?

seems to me that in a free fall if the apple and myself were pulled into one another over a long time it would be purly coincidental.

I guess i am sort of changing my question by introducing long intervals of time...

5. May 1, 2013

### Staff: Mentor

In a universe with only you and the apple, the apple could perform a very slow orbit around you, indeed (something like 1 revolution per day). If it goes to fast, it will escape and you'll never see it again.

6. May 1, 2013

### Paintjunkie

situation 1) only me and apple - we orbit each other over time
situation 2) me apple and earth - me and apple orbit earth, but not each other

observed over time each situation will produce very different observations. so how is it that people can say free fall is like being weightless because of inertial frames or w/e.

7. May 1, 2013

### Travis_King

In orbit, we are not really that high up. We are still well within Earth's sphere of influence and, if I remember correctly, at the distance that the ISS orbits we are still experiencing something like 90% of the surface gravity. However, weight is a force (more specifically, a contact force). On Earth we feel heavy because the ground is exerting a force which is equal and opposite to the force exerted by Earth's gravitational acceleration (multiplied by our mass). When we are in orbit, however, we are in a free fall. There are forces acting on us (gravity), but that force goes into accelerating us (hence, the fall). There is nothing pushing against us, resisting the gravitational force, and therefore we are "weightless".

If you free fall in an elevator, you will feel weightless because the floor will be falling away from you at the same rate that you are being pushed into it. No force will be exerted on you by the elevator, and hence you will feel weightless. You still have mass. You are still accelerating at -9.81 m/s^2. But you have no weight because nothing is resisting your fall.

When you talk about actual circumstances (i.e. you and an apple in a space station) versus theoretical circumstances (i.e. you and an apple and that is it), you can get confused...

The force between the apple and you is extremely, extremely low, especially compared to the force between the apple and the Earth.

Edit: Also, you have the idea about "orbits" slightly wrong. Not everything will necessarily orbit. You need certain conditions of distance and speed in order to create a proper orbit. As mfb said, if the thing's got too much velocity, it will simply sail away into the abyss for eternity. If it's got too little velociy, it'll smack into you over time.

Last edited: May 1, 2013
8. May 1, 2013

### xAxis

Also, there is no difference between your two cases. If the apple orbits you in free space, it will orbit you even if you orbit a planet, given the same distance and relative speeds between you and an apple.

9. May 1, 2013

### Staff: Mentor

The apple will orbit you in both situations, it's really no different than the way the space station orbits the earth while the earth orbits the sun. However, the gravitational force between you and the apple is so weak and hence the orbital velocity of the apple around you is so small, that we don't really notice. Round numbers, if I haven't messed up doing the math in my head, an apple one meter from you will orbit you with a speed of a centimeter an hour or so, one orbit per month or thereabouts.

10. May 2, 2013

### Paintjunkie

thank you everyone. I will revise my thinking

11. May 2, 2013

### sophiecentaur

Actually, you could 'easily' arrange for a small ball bearing to close-orbit a steel* sphere of a metre radius and it would have a period of around 90minutes - same as a low Earth orbit.
I would love for someone to do that. Problem is that real spacecraft have a density as low as possible so the period of your apple would be months / years?. But satellites do actually interact and would mutually orbit if they were not kept on station - as do the bits of debris in Saturn's rings and the asteroid belt..

* or with a density equivalent to that of the Earth's average

12. May 2, 2013

### Staff: Mentor

A low earth orbit does not allow orbits of smaller objects around you. The problem is the inhomogeneous gravitational field of earth: it is not the same where you are and where the apple is. This effect is small within the scale of a human, but the mass of a human is small, too (here is the right place to insert jokes about mothers).
To get enough gravitational force to counter that, you would need a density of more than 100 times the density of lead. Alternatively, go into a higher orbit. It is no problem to orbit the moon, for example.
We even have this effect on larger scales: Moon is orbiting earth, while the system "earth+moon" orbits the sun.

13. May 3, 2013

### sophiecentaur

Interesting comment and it makes sense qualitatively. I assume the inhomogeneity you refer to due to ISL and not the mass distribution within the Earth? I can see that the variation of field due to ISL would depend upon the mini-orbit diameter and the diameter (and density) would give the highest possible mass of the satellite, which would give the local field due to the satellite. So are you implying that the satellite and its mini satellite would still be mutually attracted but that any orbit would not be stable?

The perturbation due to the earth would decrease as the gradient of its field ( an inverse cube law?) so, in a high orbit (GSO), the mini orbit would suffer less by a factor of at least three orders of magnitude (?) compared with LEO. By the time you get to the Moon, there is nearly a further three orders of magnitude reduction.

14. May 3, 2013

### Staff: Mentor

What is ISL?

The inhomogeneity is simply the radial field of earth - it gets weaker with increasing distance, and changes direction if you go in the horizontal direction.
It is not just stability, you do not get orbits at all - any system of masses would drift away as their orbits around earth are too different.
The Hill sphere gives a way to analyze this quantitatively. With conventional materials (density below 25g/cm^2), you have to go beyond the level of geostationary orbits to get orbits around your orbiting object.
Right.

15. May 4, 2013

### Popper

Let's consider two cases

Case #1 - You're in free-fall in a uniform gravitational field near an apple that is initially at rest relative to you.

Case #2 - You're in an inertial frame of reference which is far removed from any gravitational source (and thus in flat spacetime).

In case #1, and as measured in your free-fall frame, the apple will accelerate towards you at a finite (although small) rate of acceleration, call it a.

In case #1, and as measured in your inertial frame, the apple will accelerate towards you at a finite (although small) rate of acceleration, call it b.

Then if can be shown that a = b. Thus you cannot tell whether you're in free fall in a uniform gravitational field or in an inertial frame of reference in flat spacetime. The weak equivalence principle states that a uniform gravitational field is equivalent to a uniformly accelerating frame of reference. Of course this assumes that you're in a closed lab and not observing what is going on outside the lab and nothing can get inside the lab from the outside.

If the gravitational field is non-uniform, i.e. there are tidal gradients present in the field, then you'll be able to determine if you're in an inertial frame of reference in flat spacetime or in a non-uniform gravitational field so long as you either wait long enough to observe what's going on in the lab or if the lab is large enough so that you can oberver tidal accelerations of objects placed in your region of space. This is equivalent to saying that you're observing what's going on in a large enough region of spacetime. But if you have sensitive enough instruments you can always determine if you're in a non-uniform gravitational field.

Another phrasing of the equivalence principle states that a free-fall frame is locally equivalent to an inertial frame of reference.

The strong equivalence principle states that the laws of physics are the same in all coordinate systems/frames of reference. This requires that the laws of physics be stated in covariant form which means that they are expressed using tensors (and derivatives are covariant derivatives). There is a coordinte free way in which the laws of physics can be stated and that's using the geometric version of the law which uses tensors in their geometric form rather than their component form.

16. May 4, 2013

### sophiecentaur

ISL is Inverse Square Law. (I should have put it in full)
Thanks for that link to the Hill Sphere. That picture says it all and the graph for the planets is informative.
This puts me in mind of the threads I've read on the Horseshoe Orbit.